How do you divide $\frac{2 x^{4} + 4 x + 1}{2 x + 3}$ $(2x^4+4x+1)/(2x+3)$ ?

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How do you divide $\frac{2 x^{4} + 4 x + 1}{2 x + 3}$ $(2x^4+4x+1)/(2x+3)$ ?

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Answer 1 · 2015-12-15T10:41:19

$x^{3} - \frac{3}{2} x^{2} + \frac{9}{4} x - \frac{11}{8}$ $x^3 - 3/2 x^2 + 9/4 x - 11/8$ with remainder $\frac{41}{8}$ $41/8$

Explanation:

I know that there are in some countries, a different notation of long polynomial division is being used. Let me use the notation that I'm most familiar with, I hope that it will be no problem for you to convert it into your prefered notation.

$ξ i (2 x^{4} \times \times \times \times \times x + 4 x + 1) \div (2 x + 3) = x^{3} - \frac{3}{2} x^{2} + \frac{9}{4} x - \frac{11}{8}$ $color(white)(xii)(2x^4 color(white)(xxxxxxxxxxx)+ 4x + 1) -: (2x + 3) = x^3 - 3/2 x^2 + 9/4 x - 11/8$
$- (2 x^{4} + 3 x^{3})$ $- (2x^4+ 3x^3)$
$\times \frac{\times \times \times x}{}$ $color(white)(xx)(color(white)(xxxxxxx))/()$
$\times \times x - 3 x^{3}$ $color(white)(xxxxx)- 3x^3$
$\times - (- 3 x^{3} - \frac{9}{2} x^{2})$ $color(white)(xx) - (-3x^3 - 9/2 x^2)$
$\times \times \frac{\times \times \times \times \times x}{}$ $color(white)(xxxx)(color(white)(xxxxxxxxxxx))/()$
$\times \times \times \times \times \times \frac{9}{2} x^{2} + 4 x$ $color(white)(xxxxxxxxxxxx)9/2 x^2 + 4x$
$\times \times \times \times x - (\frac{9}{2} x^{2} + \frac{27}{4} x)$ $color(white)(xxxxxxxxx)-(9/2 x^2 + 27/4 x)$
$\times \times \times \times \times x \frac{\times \times \times \times \times}{}$ $color(white)(xxxxxxxxxxx)(color(white)(xxxxxxxxxx))/()$
$\times \times \times \times \times \times \times x - \frac{11}{4} x + i 1$ $color(white)(xxxxxxxxxxxxxxx) -11/4 x + color(white)(i) 1$
$\times \times \times \times \times \times - (- \frac{11}{4} x - \frac{33}{8})$ $color(white)(xxxxxxxxxxxx) -(-11/4 x -33/8)$
$\times \times \times \times \times \times \times \frac{\times \times \times \times \times x}{}$ $color(white)(xxxxxxxxxxxxxx)(color(white)(xxxxxxxxxxx))/()$
$\times \times \times \times \times \times \times \times \times \times \times x \frac{41}{8}$ $color(white)(xxxxxxxxxxxxxxxxxxxxxxx)41/8$

Thus, your quotient is

$x^{3} - \frac{3}{2} x^{2} + \frac{9}{4} x - \frac{11}{8}$ $x^3 - 3/2 x^2 + 9/4 x - 11/8$

and your remainder is $\frac{41}{8}$ $41/8$ .

In total,

$\frac{x^{4} + 4 x + 1}{2 x + 3} = x^{3} - \frac{3}{2} x^{2} + \frac{9}{4} x - \frac{11}{8} + \frac{41}{8 (2 x + 3)}$ $(x^4 + 4x + 1)/(2x + 3) = x^3 - 3/2 x^2 + 9/4 x - 11/8 + 41/(8(2x+3))$

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How do you divide $\frac{2 x^{4} + 4 x + 1}{2 x + 3}$ $(2x^4+4x+1)/(2x+3)$ ?

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How do you divide $\frac{2 x^{4} + 4 x + 1}{2 x + 3}$ $(2x^4+4x+1)/(2x+3)$ ?